A cuboid measures 2 × 3 × 5 . Find the volume of revolution that results when the cuboid is rotated about its space diagonal (a space diagonal is a line segment connecting two opposite vertices such that the center of the cuboid bisects it). The use of a programming environment is highly recommended.
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Excellent solution. Thanks for sharing it.
A fun little project indeed!
Let the space diagonal be between points ( 0 , 0 , 0 ) and ( 2 , 3 , 5 ) . Let u be a unit vector along the space diagonal, and let u 1 and u 2 be unit vectors perpendicular to u .
Let the parameter α be the distance along the space diagonal. Let the value σ be the perpendicular distance from a point on the space diagonal to some place on the cuboid. Let θ be an angular parameter.
u 3 = c o s θ u 1 + s i n θ u 2 Point on Cuboid Surface = α u + σ u 3
Now, the tricky part is that for each α , θ varies continuously between 0 and 2 π . Consequently, there are infinitely many σ values for each α . The processing is as follows:
1)
For every
α
between
0
and the length of the space diagonal (in small steps), sweep
θ
from
0
to
2
π
in small steps.
2)
For every
θ
, form
u
3
and determine a
σ
value corresponding to the intersection with the infinite plane containing each cuboid face. There will be six
σ
calculations in total.
3)
The intersection point with the cuboid corresponds to the smallest positive
σ
value. Call this
σ
i
n
t
e
r
s
e
c
t
4)
For each
α
, as
θ
is swept, keep track of the largest value of
σ
i
n
t
e
r
s
e
c
t
. Call this
σ
m
a
x
5)
This yields a disk of area
π
σ
m
a
x
2
for each
α
6)
The infinitesimal volume associated with each
α
value is
π
σ
m
a
x
2
d
α
7)
Summing up all of the infinitesimal volumes yields a total volume of rotation of about
9
6
.
5
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Excellent solution. Thank you for sharing it.
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Using the following code, the result for a = 2 , b = 3 , and c = 5 is V ≈ 9 6 . 6 1 3 7 .